Proof of The Convex Hull Consists of the Convex Combinations of Finitely Many Points

Let $K$ denote the set on the right-hand side of the display. Recall from Convex Hull of a Subset of $\mathbb{R}^n$ (i) that $\operatorname{conv}S$ is convex, contains $S$, and is contained in every convex subset of $\mathbb{R}^{n}$ that contains $S$. We prove the two inclusions $\operatorname{conv}S\subseteq K$ and $K\subseteq\operatorname{conv}S$.

For the first inclusion it suffices, by the minimality of the convex hull, to show that $K$ is a convex set containing $S$. Every $x\in S$ lies in $K$: take $N=1$, $x_{1}=x$ and $t=(1)$, which lies in $\Delta_{1}$, and then $\sum_{k=1}^{1}t_{k}x_{k}=x$ by Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (i). To see that $K$ is convex, let $u,v\in K$ and $\lambda\in[0,1]$, and write

$$u=\sum_{k=1}^{N}\tau_{k}x_{k},\qquad v=\sum_{j=1}^{P}\sigma_{j}x'_{j}$$

with $N,P\in\mathbb{N}$, points $x_{1},\dots,x_{N},x'_{1},\dots,x'_{P}\in S$, and weights $\tau\in\Delta_{N}$, $\sigma\in\Delta_{P}$. Let $w_{1},\dots,w_{N+P}$ be the concatenated points ($w_{k}=x_{k}$ for $k\in[N]$, $w_{N+j}=x'_{j}$ for $j\in[P]$) and $t\in\mathbb{R}^{N+P}$ the combined weights ($t_{k}=(1-\lambda)\tau_{k}$, $t_{N+j}=\lambda\sigma_{j}$). By Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (v), $t\in\Delta_{N+P}$ and

$$(1-\lambda)u+\lambda v=\sum_{k=1}^{N+P}t_{k}w_{k}.$$

Every $w_{k}$ lies in $S$, so $(1-\lambda)u+\lambda v\in K$. Thus $K$ contains the segment $[u,v]$ of Segments and Convex Subsets of $\mathbb{R}^n$ (i) whenever $u,v\in K$, that is, $K$ is convex in the sense of Segments and Convex Subsets of $\mathbb{R}^n$ (ii). Hence $\operatorname{conv}S\subseteq K$.

For the reverse inclusion, let $N\in\mathbb{N}$, $x_{1},\dots,x_{N}\in S$ and $t\in\Delta_{N}$. The set $\operatorname{conv}S$ is convex and contains each $x_{k}$, so it contains $\sum_{k=1}^{N}t_{k}x_{k}$ by Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (vi). Hence $K\subseteq\operatorname{conv}S$, and the two sets are equal.